Question

How many terms to be added of an AP 2+5+8+---to make the
Sum 610
​

Answer 1

Answer:

2+5+8=15

610-15 =595

so 595 should be added

Answer 2

Answer:

20 terms

Step-by-step explanation:

given, sn = 610

a1 = 2

d = 5-2 =3

sn = n/2 [2a1 + (n-1) d]

610 = n/2 [ 2×2 + (n-1) 3]

610 ×2 = n [ 4 + (n-1)3 ]

1220 = n [ 4 + 3n -3 ]

1220 = n [1+3n]

1220 = n + 3n²

3n² + n - 1220 = 0

3n²+61n -60n - 1220 = 0

n( 3n + 61 ) -20 ( 3n +61 ) = 0

(n-20 ) (3n+61) = 0

n= 20 or n = -61/3

negleting the negative value for n as n can only be positive,

n = 20

so, 20 terms of an AP (2+5+8+..) make the sum 610.

check =>

a20 = 2 + 19×3

a20 = 59

610 = 20/2 ( A1+A20)

610 = 10 ( 2+59)

610 = 10× 61